题目：
| MPI Maelstrom| Time Limit: 1000MS | | Memory Limit: 10000K |
| --------------------------- | - | ------------------------ |
| Total Submissions: 9059 | | Accepted: 5565 |DescriptionBIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has asked her to benchmark the new system. ``Since the Apollo is a distributed shared memory machine, memory access and communication times are not uniform,'' Valentine told Swigert. ``Communication is fast between processors that share the same memory subsystem, but it is slower between processors that are not on the same subsystem. Communication between the Apollo and machines in our lab is slower yet.'' ``How is Apollo's port of the Message Passing Interface (MPI) working out?'' Swigert asked. ``Not so well,'' Valentine replied. ``To do a broadcast of a message from one processor to all the other n-1 processors, they just do a sequence of n-1 sends. That really serializes things and kills the performance.'' ``Is there anything you can do to fix that?'' ``Yes,'' smiled Valentine. ``There is. Once the first processor has sent the message to another, those two can then send messages to two other hosts at the same time. Then there will be four hosts that can send, and so on.'' ``Ah, so you can do the broadcast as a binary tree!'' ``Not really a binary tree -- there are some particular features of our network that we should exploit. The interface cards we have allow each processor to simultaneously send messages to any number of the other processors connected to it. However, the messages don't necessarily arrive at the destinations at the same time -- there is a communication cost involved. In general, we need to take into account the communication costs for each link in our network topologies and plan accordingly to minimize the total time required to do a broadcast.''InputThe input will describe the topology of a network connecting n processors. The first line of the input will be n, the number of processors, such that 1 <= n <= 100. The rest of the input defines an adjacency matrix, A. The adjacency matrix is square and of size n x n. Each of its entries will be either an integer or the character x. The value of A(i,j) indicates the expense of sending a message directly from node i to node j. A value of x for A(i,j) indicates that a message cannot be sent directly from node i to node j. Note that for a node to send a message to itself does not require network communication, so A(i,i) = 0 for 1 <= i <= n. Also, you may assume that the network is undirected (messages can go in either direction with equal overhead), so that A(i,j) = A(j,i). Thus only the entries on the (strictly) lower triangular portion of A will be supplied. The input to your program will be the lower triangular section of A. That is, the second line of input will contain one entry, A(2,1). The next line will contain two entries, A(3,1) and A(3,2), and so on.OutputYour program should output the minimum communication time required to broadcast a message from the first processor to all the other processors.Sample Input5 50 30 5 100 20 50 10 x x 10Sample Output```
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思路：
这道题的题意是给了你n个点，在进行输入的时候，以第一个样例来说，有5个顶点，顶点与顶点之间的权以一个三角形的方式来给出，因为自己到自己肯定是0，所以就没有给从自己到自己，第二行是map\[2]\[1]=50,第二行第一个意味着从1号顶点到2好顶点的权值是50，在第三行里map\[3]\[1]=30,map\[3]\[2]=5,意思是第三行的第一个和第二个,从3号顶点到1号顶点权值为30，从3号顶点到2号顶点权值为5，当输入的值为"x"，的时候证明这个点没有通路，也就是设置为无穷大，求从第一个点到任意一点的最小距离里面最大的一个。看了看网上很多博客，他们处理x的时候把字符串转换成整数的时候在自己写了函数，效率不好。其实在#include <:stdlib.h>里有一个把文本型转换成整形的函数atoi()来进行转换，具体看代码
代码：
```cpp
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define N 100+20
#define M 100000+20
#define inf 0x3f3f3f3f
using namespace std;
int map[N][N],dis[N],vis[N];
int n;
void dijkstra(int n)
{
int k,min;
for(int i=1; i<=n; i++)
{
dis[i]=map[1][i];
vis[i]=0;
}
for(int i=1; i<=n; i++)//遍历顶点
{
k=0;
min=inf;
for(int j=1; j<=n; j++)
if(!vis[j]&&dis[j]<min)
{
min=dis[j];
k=j;
}
vis[k]=1;
for(int j=1; j<=n; j++)
if(!vis[j]&&dis[k]+map[k][j]<dis[j])
dis[j]=dis[k]+map[k][j];//如果找到了通路就加上
}
return;
}
int main()
{
while(~scanf("%d",&n))
{
int maxx=0;
char s[10];
mem(map,inf);
mem(vis,0);
mem(dis,0);
for(int i=1; i<=n-1; i++)
{
for(int j=1; j<=i; j++)
{
scanf("%s",s);
if(strcmp(s,"x")==0)
map[i+1][j]=map[j][i+1]=inf;
else
map[i+1][j]=map[j][i+1]=atoi(s);
}
}
for(int i=1; i<=n; i++)
map[i][i]=0;
dijkstra(n);
for(int i=1; i<=n; i++)
if(dis[i]>maxx)
maxx=dis[i];//从所有最短路中找出最大的
printf("%d\n",maxx);
}
return 0;
}
\
